4Al(s) + 3O2(g)2Al2O3(s)

Delta H reaction = -3351.4 kJ

How much energy is transferred when 42.6 g of aluminum reacts to form aluminum oxide?

Thanks

well, four moles of Al yields 3351J

figure the moles of Al in 42.6g

moles= 42.6/atomicmassAl

molesgiven/4=Heat/3351.4kJ
solve for Heat.

moles= 42.6g/26.9

(42.6/26.9)/4=heat/3351J

To find out how much energy is transferred when 42.6 g of aluminum reacts to form aluminum oxide, we can use the concept of stoichiometry and the given enthalpy change of the reaction.

First, we need to calculate the number of moles of aluminum (Al) in 42.6 g. To do this, we use the molar mass of aluminum, which is 26.98 g/mol:

Number of moles of Al = mass of Al / molar mass of Al
= 42.6 g / 26.98 g/mol
≈ 1.58 mol

According to the balanced chemical equation, the reaction of aluminum with oxygen (O2) produces aluminum oxide (Al2O3) in a 4:2 ratio. This means that for every 4 moles of aluminum, 2 moles of aluminum oxide are formed. Therefore, we multiply the number of moles of Al by a factor of 2/4:

Number of moles of Al2O3 = (1.58 mol) x (2/4)
= 0.79 mol

Now, we can use the enthalpy change (ΔH) of the reaction to calculate the energy transferred. The given value is -3351.4 kJ for the reaction:

Energy transferred = ΔH x (number of moles of Al2O3)
= -3351.4 kJ/mol x 0.79 mol
≈ -2646.1 kJ

Therefore, approximately 2646.1 kJ of energy is transferred when 42.6 g of aluminum reacts to form aluminum oxide. The negative sign indicates that the reaction is exothermic, meaning that energy is released.

When calculating the moles of Al in 42.6 g, do you multiply the molar mass by four or just Mr of one Al atom?

Thanks